On-policy state distribution for episodic tasks on Sutton & Barto, page 199
Let's first assume that there is only one action so that $\pi(a|s) = 1$ for every state - action pair which simplifies the discussion.
Now let's consider a case with 100 time steps, 10 states and uniform distribution for starting state $s_0$ with $h(s_0) = 1$. The result would be
\begin{align}
\eta(s_0) &amp;= 1 + \sum_{i = 0}^9 \eta(s_i) \cdot p(s_0|s_i) =\\
&amp;= 1 + \sum_{i = 0}^9 10 \cdot \frac{1}{10} = 11
\end{align}
Now let's consider a case with 1000 time steps where other settings are the same as in the first case.
\begin{align}
\eta(s_0) &amp;= 1 + \sum_{i = 0}^{9} \eta(s_i) \cdot p(s_0|s_i) =\\
&amp;= 1 + \sum_{i = 0}^{9} 100 \cdot \frac{1}{10} = 101
\end{align}
In the first case
\begin{equation}
\mu(s_0) = \frac{11}{9\cdot 10 + 11} = 0.1089
\end{equation}
and in the second case you have
\begin{equation}
\mu(s_0) = \frac{101}{9\cdot 100 + 101} = 0.1009
\end{equation}
so it looks like you are correct that $\mu(s)$ depends on the length of the episode, but they didn't really say that it doesn't. Obviously as the length of the episode increases so will the number of times a certain state was visited so you could say that formula implicitly depends on the number of time steps. If $h(s_i)$ is equal for every state then results would be the same in both cases regardless of number of time steps. Also, as the number of possible states grows very large, as it usually is in real problems, the results would be approaching each other as the number of states grows.